Urea, #("NH"_2)_2"CO"#, is dissolved in #"100 g"# of water. The solution freezes at #-0.085^@"C"#. How many grams of urea were dissolved to make this solution?

Question

(Given #K_f# of water = 1.858 #K_b# = 0.512)

Naomi Aronson · Accepted Answer

#"0.27 g"#

Your tool of choice here will be the equation that allows you to calculate the freezing point depression of the solution

#color(blue)(ul(color(black)(DeltaT_"f" = i * K_f * b)))#

Here

Now, the freezing point depression tells you the difference between the freezing point of the pure solvent and the freezing point of the solution.

In other words, the freezing point depression tells you by how many degrees the freezing point of the solution decreases compared with the freezing point of the pure solvent.

You know that your solution freezes at #-0.085^@"C"#. Since pure water has a normal freezing point of #0^@"C"#, you can say that the freezing point depression will be

#DeltaT_"f" = 0^@"C" - (-0.085^@"C")#

#DeltaT_"f" = 0.085^@"C"#

This essentially means that the freezing point of the solution is #0.085^@"C"# lower than the normal freezing point of the pure solvent.

So, rearrange the equation to find the molality of the solution.

#b = (DeltaT_"f")/(i * K_f)#

Since urea is a nonelectrolyte, which means that it does not dissociate in aqueous solution, its van't Hoff factor will be equal to #1#.

This means that you have

#b = (0.085 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.858 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = "0.04575 mol kg"^(-1)#

This tells you that this solution contains #0.04575# moles of urea for every #"1 kg" = 10^3 quad "g"# of solvent. Since your sample contains #"100 g"# of water, you can say that it will also contain

#100 color(red)(cancel(color(black)("g water"))) * "0.04575 moles urea"/(10^3color(red)(cancel(color(black)("g water")))) = "0.004575 moles urea"#

Finally, to convert the number of grams to moles, use the molar mass of urea

#0.004575 color(red)(cancel(color(black)("moles urea"))) * "60.06 g"/(1color(red)(cancel(color(black)("mole urea")))) = color(darkgreen)(ul(color(black)("0.27 g")))#

I'll leave the answer rounded to two sig figs, but don't forget that you have one significant figure for the mass of water.

Cooper Adams · Answer

undefined To find the mass of urea dissolved in the solution, we can use the freezing point depression equation:

ΔT = K_f * m

Where:
ΔT = Freezing point depression (in °C)
K_f = Freezing point depression constant for water (1.86 °C/m for water)
m = molality of the solute (moles of solute per kilogram of solvent)

First, we need to find the molality of the urea solution:

ΔT = -0.085 °C
K_f = 1.86 °C/m
m = ?

ΔT = K_f * m
m = ΔT / K_f
m = (-0.085 °C) / (1.86 °C/m)
m ≈ -0.0457 mol/kg

Now, we need to convert molality to moles of urea dissolved:

1 kg of water = 1000 g of water
So, 100 g of water = 0.1 kg

moles of urea = molality * mass of water (in kg)
moles of urea = -0.0457 mol/kg * 0.1 kg
moles of urea ≈ -0.00457 mol

Since urea is non-electrolyte, it dissociates into 2 ions in water, so the molar mass of urea (60.06 g/mol) is divided by 2 to find the mass of one mole of urea:

molar mass of urea = 60.06 g/mol / 2 = 30.03 g/mol

Now, we can find the mass of urea dissolved:

mass of urea = moles of urea * molar mass of urea
mass of urea ≈ -0.00457 mol * 30.03 g/mol
mass of urea ≈ -0.1375 g

However, since mass cannot be negative, we take the absolute value of the result:

mass of urea ≈ 0.1375 g

So, approximately 0.1375 grams of urea were dissolved to make this solution.